Prove or disprove that if $f(x)=\sum_{k=0}^na_kx^k$ then $\sum_{k=0}^{n+1}(-1)^{n+1-k}\binom{n+1}kf(x+k)=0$

Question

I had the following problem:
Let $f(x)=ax^2+bx+c$. Find $f(x+3)-3f(x+2)+3f(x+1)-f(x)$.
It is very simple to solve this and the result is $0$. But, I noticed that coefficients of second equality $(1,-3,3,-1)$ is the same as coefficients of $(x-1)^3$. So, I came with the following hypothesis:

Let $f(x)=\sum_{k=0}^na_kx^k$ for all $x\in\mathbb{R}$ and some $n\in\mathbb{N},a_0,a_1,\dots,a_n\in\mathbb{R}$. Then $$\sum_{k=0}^{n+1}(-1)^{n+1-k}\binom{n+1}kf(x+k)=0$$

I tried to find some way to prove or disprove it. I firstly tried induction. It is true for $n=\{1,2,3\}$, so I supposed it is true for some $m\in\mathbb{N}$. Then I realized that it is not easy to prove it for $m+1$ because the coefficients isn't same. Also, it is very different for odd and even $n$.
My second attempt is to try to group terms in some way. So, I have tried to find $$\sum_{k=0}^{n}(-1)^{n-k}\binom nk$$ It is easy to prove that for any odd $n$ it is equal to $0$. Then I proved that for even $n$ we can group terms so we can reduce it to the first case (odd $n$), so for all $n\in\mathbb{N}$ it is equal to $0$.
Now, problem is $f(x+k)$. After many attempts, I couldn't find any way to group terms and reduce it to $0$.
What is the easiest way to prove or disprove it?

Answer 1

Algebraic method: set $E=\Bbb R_n[X]=\{P\in \Bbb R[X] / \ \deg P\leq n \}$ and $\varphi:E\to E$ defined by $\varphi(P)=P(X+1)$, then $\varphi$ is a linear map and $\varphi^k(P)=P(X+k)$. Now use the Cayley-Hamilton's Theorem.

Answer 2

One of the reasons why I've written out m.se answer #1379518 in all that detail was my expectation that it would come up over and over. Well, this is such a case. Fix $x \in \mathbb{R}$. In the notations of this answer, the sequence $\left(f\left(x+0\right),f\left(x+1\right),\cdots,f\left(x+n+1\right)\right)$ is $n$-polynomial (since $f\left(x+t\right)$ is a polynomial in $t$ of degree $\leq n$), and thus Theorem 3 in my post (applied to $\mathbf{a} = \left(f\left(x+0\right),f\left(x+1\right),\cdots,f\left(x+n+1\right)\right)$ and $k=n$) shows that the sequence $\Delta^{n+1}\left(f\left(x+0\right),f\left(x+1\right),\cdots,f\left(x+n+1\right)\right)$ consists solely of zeroes. But Theorem 2 in the same post (applied to $n+2$, $\left(f\left(x+0\right),f\left(x+1\right),\cdots,f\left(x+n+1\right)\right)$, $n+1$, and $1$ instead of $n$, $\left(a_1,a_2,\ldots,a_n\right)$, $k$ and $p$) shows that the $1$-st entry of this sequence is $\sum\limits_{i=0}^{n+1}\left(-1\right)^{n+1-i}\dbinom{n+1}{i}f\left(x+i\right)$. Thus, $\sum\limits_{i=0}^{n+1}\left(-1\right)^{n+1-i}\dbinom{n+1}{i}f\left(x+i\right) = 0$. This is your claim.

Answer 3

The claim holds if $$\sum_{k=0}^{n+1} {n+1\choose k} (-1)^k (x+k)^q = 0$$ where $0\le q\le n.$

Introduce $$(x+k)^q = \frac{q!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}} \exp((x+k)z) \; dz.$$

This yields for the sum $$\frac{q!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}} \exp(xz) \sum_{k=0}^{n+1} {n+1\choose k} (-1)^k \exp(kz) \; dz \\ = \frac{q!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{q+1}} \exp(xz) (1-\exp(z))^{n+1}.$$

This is $$q! [z^q] \exp(kz) (1-\exp(z))^{n+1}.$$

Note however that $$1-\exp(z) = -z - \frac{1}{2} z^2 -\frac{1}{6} z^3 - \cdots$$

so that $(1-\exp(z))^{n+1}$ starts at $z^{n+1}.$ Hence the coefficient on $z^q$ is zero because we assumed that $0\le q\le n.$

Answer 4

That is a well known result in the field of Finite Differences.

In fact, if we define $$ \Delta \,f(x) = f(x + 1) - f(x) $$ then its iteration is $$ \Delta ^{\,n} \,f(x) = \Delta \,\left( {\Delta ^{\,n - 1} f(x)} \right) = \sum\limits_{0\, \le \,k\, \le \,n} {\left( { - 1} \right)^{\,n - k} \left( \matrix{ n \cr k \cr} \right)f(x+k)\,} $$

It is easy to demonstrate that the $\Delta$ of a polynomial of degree $q$ provides a polynomial with one degree less.

Answer 5

Writing it as $\,f_x = \left(\sum_{k=0}^na_kx^k\right) \cdot 1^x\,$ makes $\,f_x\,$ better recognizable as a solution of the linear recurrence with constant coefficients having the characteristic polynomial with the only root $\,1\,$ of multiplicity $\,n+1\,$, which is $\,p(t)=(t-1)^{n+1} = \sum_{k=0}^{n+1}\,(-1)^{n-k+1}\binom{n+1}{k}\,t^k\,$.

Since $\,f_x\,$ satisfies the recurrence defined by $\,p(t)\,$, the sum is zero when replacing $t^k \mapsto f_{x+k}\,$.

Answer 6

In fact, (finite) induction does work on this problem but it's not immediate. There are already many good answers but for the sake of completeness, I will show how it can be done. The key is to use two representations of the polynomials and a little calculus trick that occurs frequently in such calculations. Let $n \in \mathbb{N}$, and for $p \in [0,n-1]$, let $\mathcal{H}(p):"\text{If } P \text{ is a polynomial such that} \deg P \leq p, \text{then} \sum_{k=0}^n \binom{n}{k} (-1)^kP(k)=0"$.

• $\mathcal{H}(0)$ is true by Newton's binomial formula.
• Let $p \in [0...n-2]$ such that $\mathcal{H}(p)$ is true. Let us show that $\mathcal{H}(p+1)$ is true. Let $P$ be a polynomial of degree $p+1$ .

We have $$P=\sum_{i=0}^{p+1}a_iX^i$$for some collection $(a_i)$ of complex numbers.Since $Q=\sum_{i=0}^{p}a_iX^i$ is a polynomial of degree at most $p$, it verifies the induction hypothesis. We then have $$\sum_{k=0}^n \binom{n}{k} (-1)^k\left[Q(k)+a_{p+1} k^{p+1}\right]=\underbrace{0}_{\text{I.H.}}+\sum_{k=0}^n \binom{n}{k}(-1)^k a_{p+1} k^{p+1}$$ It is therefore enough to show that $$\sum_{k=0}^n \binom{n}{k}(-1)^k k^{p+1}=0$$ To this end, consider the function $f:x \mapsto (1+x)^n.$ By Newton's binomial, we have $\forall x \in \mathbb{R},f(x)=\sum_{k=0}^n \binom{n}{k}x^k$. Besides, $f$ can be differentiated over $\mathbb{R}$ and its derivatives also have (at least) two expressions each: $$\forall j \in \mathbb{N}, f^{(j)}:x\mapsto \begin{equation} X= \begin{cases} 0 & \text{if}\ j>n \\ \frac{n!}{(n-j)!}(1+x)^{n-j}=\sum_{k=j}^{n} \binom {n} {k} \left(\prod_{i=0}^{j}(k-i)\right)x^{k-j} & \text{otherwise} \end{cases} \end{equation}$$ Notice how the product on the left hand side was simplified using factorials whereas the one on the right was kept. In particular, if we let $$A = \underbrace{\prod_{i=0}^{p}(X-i)}_{\text{Polynomial of degree p+1}} = \sum_{i=0}^{p+1} \alpha_kX^k$$ where $(\alpha_k)$ are complex numbers such that $\alpha_{p+1}=1$ then we have $$f^{(p+1)}(-1)=\sum_{k=p+1}^{n} \binom {n} {k}(-1)^{k-p-1}A(k)$$ As $p+1<n$, we have $f^{(p+1)}(-1)=0$, so $$\sum_{k=p+1}^{n} \binom {n} {k}(-1)^{k-p-1}A(k)=0$$ However, by definition of $A$, $\forall k \in [0,p] A(k)=0$ so $$\sum_{k=0}^{n} \binom {n} {k}(-1)^kA(k)=0$$ By using the induction hypothesis again, as well as linearity, we obtain $$\underbrace{\alpha_{p+1}}_{=1}\sum_{k=0}^{n} \binom {n} {k}(-1)^k k^{p+1}=0$$ as desired.

Side note: I found this elementary solution while working on homework where we had to prove results on finite recurrence relations, so in my case using them as it was done in another answer would be circular reasoning. The derivative trick seems linked to "umbral calculus" but I admittedly don't know enough on the topic to make that clear.

See also : How to compute $\sum^n_{k=0}(-1)^k\binom{n}{k}k^n$ and Orthogonality for Binomial Coefficients